## Integrable solutions to an elliptic PDE on divergence form have a definite sign

Let $f\colon\mathbb{R}^n\to\mathbb{R}^n$ be a smooth, bounded vector field. Further, let $u\colon\mathbb{R}^n\to\mathbb{R}$ satisfy $$-\Delta u=\operatorname{div}(fu).$$ If $u\in L^1(\mathbb{R}^n)$, then $u$ has one sign, i.e., either $u>0$ everywhere, or $u<0$ everywhere, or $u=0$ everywhere.

I have a direct proof of this for $n=1$. For $n>1$, I have a proof using the theory of parabolic equations (see below). My question: Is there a direct proof using only the theory of elliptic PDEs?

(Edited to assume $f$ bounded and fix the case $n=1$ below.)

For $n=1$, my proof goes as follows. The equation is $-u''=(fu)'$, which integrates to $u'+fu=A$ for some constant $A$. If $u$ changes sign then we may without loss of generality take $u(0)=0$. Thus $$u(x)=A\int_0^x e^{F(t)-F(x)}\,dt$$ where $F'=f$. If $|f|\le c$ then $F(t)-F(x)\ge c(t-x)$ for $t<x$, so $u(x)\ge Ac^{-1}(1-e^{-cx})$ when $x>0$, and hence $u\notin L^1$ (unless $A=0$).

For $n>1$, my only proof is much more involved. Here is a brief outline. Assume the conclusion is wrong, so we can write $u=u_+-u_-$ with $u_\pm\ge0$ everywhere and neither identically zero, and $u_+u_-=0$ everywhere.

Now let $v_\pm$ solve $$\frac{\partial v_\pm}{\partial t}=\Delta v_\pm+\operatorname{div}(fv_\pm)$$ for $t>0$, with initial conditions $v_\pm(0,x)=u_\pm(x)$. By uniqueness for this equation (with suitable growth conditions at infinity), $v_+(t,x)-v_-(t,x)=u(x)$ for $t>0$ and $x\in\mathbb{R}^n$. Also, for $t>0$ we find $v_\pm>0$ everywhere, and also $$\int_{\mathbb{R}^n} v_\pm(t,x)\,dx=\int_{\mathbb{R}^n} u_\pm(x)\,dx$$ since the equation is on divergence form. We conclude $$\int_{\mathbb{R}^n}|u(x)|\,dx=\int_{\mathbb{R}^n}(u_+(x)+u_-(x))\,dx=\int_{\mathbb{R}^n}(v_+(t,x)+v_-(t,x))\,dx>\int_{\mathbb{R}^n}|v_+(t,x)-v_-(t,x)|\,dx,$$ which is a contradiction.

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 I might add a bit of intuition for why this is true: The laplacian makes stuff diffuse, while the divergence term transports stuff along the vector field while preserving the total amount of stuff. If a solution has two signs, the diffusion will mix the positive and negative parts together, making them shrink. This intuition is of course formalized in the parabolic proof. – Harald Hanche-Olsen Feb 9 2010 at 0:52 Regarding my edit: It may be that it is sufficient to assume $f$ grows at most linearly. Some such restriction is necessary, though: I can make a counterexample for $n=1$ with $f$ odd, $f(x)=ax^{a-1}$ for $x>0$ where $a>1$. The odd solution $u$ with $A=1$ in the notation of the question will belong to $L^1$. We find $\int_0^\infty u=\int_0^\infty \int_0^x e^{y^a-x^a}\,dy\,dx$. The part where $y<1$ is no problem, and the other part is handled by noting $y^a-x^a ## 1 Answer I love this problem and have spent half the evening thinking about it. Here is a rough sketch of an idea that could possibly work. Making it rigorous might be a bit of a chore due to the unboundedness of$\mathbb{R}^n$, etc, but my guess is that it is doable. Let$L$denote the elliptic operator $Lu : = -\Delta u - \mathrm{div}( f u ).$ Suppose the operator$L$has a principal eigenvalue$\lambda_1$, which is the smallest number$\lambda$for which there exists a nonzero solution of the equation$Lu = \lambda u$in$\mathbb{R}^n$. Then$\lambda_1$should be simple (!) and have a principal eigenfunction which does not change sign. Let$\varphi \in L^1(\mathbb{R}^n)$denote the principal eigenfunction, which we normalize to be positive. Assume for now that$\varphi$and its derivatives tend to zero at infinity, and$f$and its derivatives stay bounded. Then we may simply integrate the equation$L\varphi = \lambda_1\varphi$to get $\lambda_1 \int_{\mathbb{R}^n} \varphi dx = 0$. Well, that means that$\lambda_1 = 0$. Recalling the simplicity of$\lambda_1=0$, we see that the equation$Lu = 0$not only has a positive solution, its set of solutions is precisely $\{ c \varphi : c\in \mathbb{R} \}$`. This implies the result. Now, you may have already noticed that there aren't really such principal eigenvalues in general, for example when$f=0$. But I think it is possible that the idea can still be made into a rigorous proof. If we look at a really large domain, say the ball$B(0,R)$with$R> 0$very large, there is a principal eigenvalue$\lambda_{1,R}$of$L$on$B(0,R)$and it is going to be close to zero. This can be shown by considering the adjoint operator$L^*$, which has the same principal eigenvalue$\lambda_{1,R}$, and it is easy to show that$0 < \lambda_{1,R} \leq CR^{-2}$, due to boundedness of$f$. This is where we use the special form of the equation. If we have a solution to$Lu = 0$on the whole space$\mathbb{R}^n$, with$u(0)>0$, I believe it should be possible to prove that if we choose the normalization$\varphi_{1,R}(0) = u(0)$, then$\varphi_{1,R}$converges to$u$(at least locally uniformly) as$R\to \infty$. In particular,$u\$ must be positive everywhere.

There are obviously some details left to work out. It is very possible I am making a silly error and it doesn't work at all.

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